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I am reading the book Lecture on mean curvature flow by Xi-Ping Zhu.

Suppose $M^n$ is an n-dimension smooth manifold and $X(x,t):M^n \rightarrow R^{n+1}$ be a one-parameter family of smooth immersion. Metric and the second fundamental form on $X(x,t)$ is defined as $$g_{ij} = (\frac{\partial X }{\partial x_i},\frac{\partial X }{\partial x_j})\,\,\,\,h_{ij}= (n(x,t),\frac{\partial^2 X }{\partial x_i\partial x_j})$$ And the covariant derivative of vector $v$ is defined by $$\nabla_j v_i = \frac{\partial v_i }{\partial x_j} + \Gamma^i_{jk}v_k$$ By setting $$\frac{\partial X }{\partial t}=H n$$ Using the Gauss equation and the Weingarten equation, the author further calculate that $$g^{ij}\nabla_i\nabla_j X=g^{ij}(\frac{\partial^2 X }{\partial x_i\partial x_j}-\Gamma_{ij}^k\frac{\partial X }{\partial x_k})=g^{ij}h_{ij}n=Hn=\frac{\partial X }{\partial t}$$ Later he used the De Turck trick to make this equation become strictly parabolic. But I think $$\frac{\partial X }{\partial t}=g^{ij}(\frac{\partial^2 X }{\partial x_i\partial x_j}-\Gamma_{ij}^k\frac{\partial X }{\partial x_k})$$ is already strictly parabolic. what is the purpose of the trick? Thank you so much.

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  • $\begingroup$ You may be careful about degenerate system and non-degenerate system $\endgroup$
    – user21574
    Commented May 1, 2017 at 7:19
  • $\begingroup$ You mean that the coefficients may be zero at some time? $\endgroup$
    – mnmn1993
    Commented May 1, 2017 at 7:22
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    $\begingroup$ I think there's a calculation in the book which expand $\Gamma_{ij}^k$ and you will also see second order derivative of $X$ from the $\Gamma_{ij}^k$ term. $\endgroup$ Commented May 2, 2017 at 6:47
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    $\begingroup$ I got what you mean @JohnMa ,thanks $\endgroup$
    – mnmn1993
    Commented May 2, 2017 at 8:13
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    $\begingroup$ Do I have to expand $\overline{\Gamma^k_{ij}}$ again to see the equation given by the trick is really strictly parabolic? $\endgroup$
    – mnmn1993
    Commented May 2, 2017 at 8:29

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Note that if $p\in M$ the symbol on some $\xi$ direction is given by $\sigma(\xi,p) = g^{ij}\xi_i\xi_j - g^{ij}\Gamma_{ij}^k\xi_k = |\xi|^2_{g} - g^{ij}\Gamma_{ij}^k\xi_k,$ and if you choose $g^{ij}\Gamma_{ij}^k\xi_k = |\xi|_g^2,$ the symbol is not an isomorphism.

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